 QUESTION

# The value of $\sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {100^ \circ }$ is equal to${\text{A}}{\text{. }}\dfrac{3}{4}$${\text{B}}{\text{. }}\dfrac{1}{8}$${\text{C}}{\text{. }}\dfrac{3}{2}$${\text{D}}{\text{. }}\dfrac{3}{{16}}$

Hint: Assume the given equation as, S = $\sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {100^ \circ }$ and use the product identity, i.e. $2\sin x\sin y = \cos (x - y) - \cos (x + y)$, and then solve the question.

Let us assume, $S = \sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {100^ \circ }$.
$S = \dfrac{1}{2}(\sin {60^ \circ }\sin {20^ \circ })(2\sin {100^ \circ }\sin {40^ \circ })$
Using the product identity, $2\sin x\sin y = \cos (x - y) - \cos (x + y)$ for $x = {100^ \circ },y = {40^ \circ }$.
$\therefore S = \dfrac{1}{2}(\sin {60^ \circ }\sin {20^ \circ })(\cos ({100^ \circ } - {40^ \circ }) - \cos ({100^ \circ } + {40^ \circ })) \\ = \dfrac{1}{2}\left( {\dfrac{{\sqrt 3 }}{2}\sin {{20}^ \circ }} \right)\{ \cos {60^ \circ } - \cos {140^ \circ }\} \\ = \dfrac{{\sqrt 3 }}{4}\sin {20^ \circ }\left( {\dfrac{1}{2} - \cos {{140}^ \circ }} \right) \\ = \dfrac{{\sqrt 3 }}{4}\left( {\dfrac{1}{2}\sin {{20}^ \circ } - \sin {{20}^ \circ }\cos {{140}^ \circ }} \right) - (1) \\$
$S = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - 2\sin {{20}^ \circ }\cos {{140}^ \circ }} \right)$
Using the product identity, $2\sin x\cos y = \sin (x + y) + \sin (x - y)$ for $x = {20^ \circ },y = {140^ \circ }$.
$\therefore S = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{20}^ \circ } + {{140}^ \circ }) - \sin ({{20}^ \circ } - {{140}^ \circ })} \right) \\ = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{160}^ \circ }) + \sin ({{120}^ \circ })} \right) \\ = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{180}^ \circ } - {{20}^ \circ }) + \sin ({{120}^ \circ })} \right) \\ = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{20}^ \circ }) + \dfrac{{\sqrt 3 }}{2}} \right) \\ = \dfrac{3}{{16}} \\$